11848 along the X-axis. What I do not get is this: How can I find the dynamic range (or SNR) you will get if I set the camera to an ISO of 25600?

If you expose up to the clipping point of the RAW file, the DR of the image would be the value indicated in the graph (6.61 EV). If you expose less than that it would be correspondingly less (one EV less DR for each EV less exposure).

The difference between the E-M5 with a camera ISO of 25600 and a measured ISO of 11848 and another camera, X, with a camera ISO of 25600 and a measured ISO of 25600 is that at the same exposure (and the meter of the two cameras would normally suggest the same exposure) the ADU values in the RAWs of camera X would be slightly more than one EV closer to the clipping point than that of the E-M5.

Consequently, when camera X is at the clipping point of the sensor, the E-M5 is slightly more than one EV below the clipping point. When the E-M5 is at the clipping point, camera X is has slightly more than one EV worth of clipping.

Thanks,

I think I understand more now. Let me try to use this in a couple of examples. _IF_ I assume that you do not want or need the extra headroom that the Olympus E-M5 provides. I then take that you can disregard the extra dynamic range above the clipping point of a "vanilla" camera where ISO 12800 actually is measured to ISO 12800.

Well, as already indicated, if you shoot the E-M5 alongside a camera X whose camera ISOs equal its measured ISOs, and camera X happens to be just at the clipping point, the below (with corrections as indicated) is what you'll end up with. And while asking yourself the questions you do is perfectly all right, I think that from a practical point of view, it makes sense to follow the advice you already got from texinwien in his latest reply to you.

Not quite. You can't insert the ISOs in the SNR formula in the manner you do. Provided that the sensor response is linear and fixed pattern noise can be ignored (which might be somewhat questionable assumptions in at least some cases, though perhaps not so much at the ISOs at issue here), the SNR increases by about 3 dB (about 3.01 to be somewhat more precise) when we increase exposure by one EV. So we have

20.9 - log2(12800/5953)*3.01 = 20.9 - 13.64*3.01 + 12.54*3.01 = 17.59

Again, please don't take this as an attempt to bash olympus or anything. I am simply trying to learn something here.